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Homework Help: Complex exponentials and differential equations

  1. Oct 23, 2005 #1
    guestion: Use complex exponentials to find the solution of the differential equation

    (d^2y(t)/dt^2) + (3dy(t)/dt) + (25/4)y(t) = 0

    such that y(0) = 0, dy/dt =1 for t=o

    my taughts: I started by putting it in the form m^2 + 3m +25/4
    m = (-3sqrt(9-25))/2 = (-3sqrt(-16))/2 = (-3+-4i)/2

    then i thaught one can put it in the form e^pt(AcosQt+BsinQt) [p+-Qi]

    so: y = (e^(-3/2)t)(Acos2t + Bsin2t)

    y(0)=0 dy/dt=1 for t=0 y=Ae^(((-3+4i)/2)t) + Be^(((-3-4i)/2)t)

    0 = (e^(-3/2)t)(Acos2t + Bsin2t)
    0 = (Acos2t + Bsin2t)
    0 = A + 0
    dy/dt = (-3/2(e^(-3/2)t))(Acos2theta + Bsin2theta) + (e^(-3/2)t)(-2Asin2t + 2Bcos2t)
    1=(-3/2)A +2B
    2B = 1 (because A=0)
    so y(t) = e^((-3/2)t) ((1/2)sin2t)

    Don't know if I have done the question right or even got the question at all. just wanted to know if this is right, or if i'm on the right track but made a misstake on the way. Also if I'm completly wrong please point that out and give me a pointer where to start. Cheers
  2. jcsd
  3. Oct 23, 2005 #2


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    It's not at all clear what you did! You apparently wrote the solution to the differential equation in terms of real functions, sine and cosine, converted those to complex exponential, found the constants A and B to give the initial conditions and then converted back to real functions. I don't believe that was what was intended by "use complex exponentials to find the solution".

    First solve the characteristic equation m2 + 3m +25/4= 0.
    (I don't like your phrasing "putting it in the form". You didn't put the differential equation in this form- this is the characteristic equation for the differential equation. Also you didn't write it as an equation, you just wrote "m2 + 3m +25/4" and then declared what m must be!)
    You solved that correctly: m= (3/2)+ i and (3/2)- i.
    Remember that one way of getting the characteristic equation is to look for a solution of the form emx. You don't need to go through the "sine", "cosine" form: the general solution to the differential equation, in terms of complex exponentials is:
    [tex]y= C_1e^{(\frac{3}{2}+ i)t}+ C_2e^{(\frac{3}{2}-i)t}[/tex]
    Taking t= 0 in that gives [tex]y(0)= C_1+ C_2= 0[/tex]
    Differentiating the general solution gives
    [tex]y'= (\frac{3}{2}+ i)C_1e^{(\frac{3}{2}+ i)t}+(\frac{3}{2}- i)C_2e^{(\frac{3}{2}+ i)t}[/tex]
    [tex]y'(0)= (\frac{3}{2}+ i)C_1+ (\frac{3}{2}+ i)C_2= 1[/tex].

    Once you have found C1 and C2 I see no reason why you can't just leave the solution in "complex exponential" form!
    Last edited by a moderator: Oct 23, 2005
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